- #1
dumbQuestion
- 125
- 0
Hello,
I understand that if we have three functions f, g, and h, they are linearly independent <=> the only c1, c2, and c3 that satisfy (c1)f+(c2)g+(c3)h=0 are c1=c2=c3=0.
In order to solve for these c1, c2, and c3, we want three equations in the three unknowns. To do this we can differentiate f, g, and h twice and construct the Wronskian. Since this is a square matrix, if the det(W =/= 0, then we know that this system is nonsingular, consistent, and the solution is unique. Furthermore, since its homogeneous we know that unique solution must be c1=c2=c3=0. So if this is the result, we know f,g, and h are linearly independent. But that also means that f', g' and h' are linearly independent, and f'', g'', and h'' are linearly independent, right?
I guess my confusion is, what if there are functions f,g, and h such that f, g, and h are linearly independent but say, f'', g'' and h'' are linearly dependent? Wouldn't this mean if we construct the Wronskian it will end up inconsistent even though f, h, and h are linearly independent? Is it even possible for that to happen?
Sorry if the question is confusing.
I understand that if we have three functions f, g, and h, they are linearly independent <=> the only c1, c2, and c3 that satisfy (c1)f+(c2)g+(c3)h=0 are c1=c2=c3=0.
In order to solve for these c1, c2, and c3, we want three equations in the three unknowns. To do this we can differentiate f, g, and h twice and construct the Wronskian. Since this is a square matrix, if the det(W =/= 0, then we know that this system is nonsingular, consistent, and the solution is unique. Furthermore, since its homogeneous we know that unique solution must be c1=c2=c3=0. So if this is the result, we know f,g, and h are linearly independent. But that also means that f', g' and h' are linearly independent, and f'', g'', and h'' are linearly independent, right?
I guess my confusion is, what if there are functions f,g, and h such that f, g, and h are linearly independent but say, f'', g'' and h'' are linearly dependent? Wouldn't this mean if we construct the Wronskian it will end up inconsistent even though f, h, and h are linearly independent? Is it even possible for that to happen?
Sorry if the question is confusing.